I am trying to power my Nikon DSLR with a permanent power source instead of the battery for long captures (overnight) without buying the overpriced battery adapter. I can't simply connect a AC-DC converter at correct voltage, as Nikon verifies the battery (or power source) and refuses to operate, if the power source is not verified. I have asked previously about a simple hack, which you can see here: simply connecting charger, battery and camera all together

This turns out to be bad, as charger's board might not handle the current spikes and the connection (charging? or not?). So the charger may be switched with a AC-DC converter. However, Nikon verifies the battery. I listened to the serial communication with a logic analyzer, which includes different data each time. I can't really decode it and if it is random, then cryptography is involved. More onto that you can find here. They even put FPGA chip inside the battery to handle verification (according to some blogs). I can't bypass the battery. However, I thought, I can use a bidirectional optocoupler to connect only the serial pin of the battery to the camera. That way, the voltage difference won't be an issue between the battery and power supply (as they will be isolated with an optocoupler) but camera will be able to communicate with the battery to verify it but won't have any clue, where the power is actually coming from.

Here is an illustration of this: Bidirectional Optocoupler-battery and camera schematic

The communication happens at 33333Hz (30μs for each bit) 3.3V logic between the battery and camera. Battery voltage can range between 7.2-8.2V. Original power adapter operat The optocoupler needs to handle 33333Hz at 3.3V. Since, I am not very familiar with power electronics and don't want to kill my camera, I would like to ask you, what can go wrong with this setup (before I try and destroy my camera)? How can I do this safely?


1 Answer 1


On my Panasonic, when I need to do long time lapse, and before the special adaptor could be bought, I hacked open an existing after-market battery, and attached two fine wires (strips of foil actually) directly to the battery tabs (the flex coming out of the cells), where they attach to the little protection pcb at the front of the battery.

They brought the battery connection around to the back of the battery, and out a hole at the back, and through the (provided) hole in the battery cover. Tape the battery case up with Kapton tape, as it is very thin.

Then float the battery with a simple lm317 regulator. The float only has to meet the average long run current drain, as the battery powers the camera during shots.


simulate this circuit – Schematic created using CircuitLab

The arrangement works fine - well it did once I thought to put the diode in...

You need to ensure you do not overcharge. If the cell voltage is 3.8V/cell (Panasonic has a 2 cell battery = 7.6V) when floating on the external power you are well away from overcharging (>4.2V/cell). This voltage keeps the battery at ~3/4 charge, which is fine for me.

Give some consideration to how you could cock things up. Note D2 to protect against reverse polarity. If you are doing this in a car, consider if any exposed metal parts of the camera are connected to one or other of the battery terminals. (if the metal was connected to B+, touching camera to car would be destructive)

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    \$\begingroup\$ So it is similar to my previous question? I can "charge" (not really) and power the camera at the same time? In your schematic, the camera would be further on the right side connected to the battery right? \$\endgroup\$ Commented Jul 3, 2018 at 21:50
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    \$\begingroup\$ Yes, it is no problem. Inside the battery is a small pcb, that disconnects the camera when the battery goes flatish (<3V/cell). If you totally flatten a lithium cell, it is destroyed. (and yes it is - I left off the diode, and R1 flattened the battery). You could also connect to the outside of the battery - the camera +/- terminals, and it would also work. Mechanically, cutting the battery open was easiest for my camera, and I chose to connect to the cells, as I won't have any unknown effect from the protection circuit that way. \$\endgroup\$
    – Henry Crun
    Commented Jul 3, 2018 at 22:08

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